본인 풀이

BFS로 풀었고 거의 모범답안에 유사하여 굳이 답안은 추가하지 않았음.

#include<bits/stdc++.h>
using namespace std;

string a[100];
int b[100][100];
int visited[100][100];
int n, m, y, x;
int dy[4] = {-1, 0, 1, 0};
int dx[4] = {0, 1, 0, -1};

int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    cin >> n >> m;

    for (int i = 0; i < n; i++){
        cin >> a[i];

        for (int j = 0; j < a[i].size(); j++){
            b[i][j] = a[i][j] - '0';
        }
    }

    queue<pair<int, int>> q;
    q.push({0, 0});
    visited[0][0] = 1;
    while (q.size()){
        tie(y, x) = q.front();
        q.pop();

        for (int i = 0; i < 4; i++){
            int ny = y + dy[i];
            int nx = x + dx[i];

            if(ny < 0 || nx < 0 || ny >= n || nx >= m){
                continue;
            }


            if(b[ny][nx] && visited[ny][nx] == 0){
                visited[ny][nx] = visited[y][x] + 1;
                q.push({ny, nx});
            }
        }
    }

    cout << visited[n - 1][m - 1] << "\n";

    return 0;
}